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bezimeni [28]
3 years ago
12

You can increase the capacitance of a capacitor by A. Decreasing the plate spacing B. Increasing the plate spacing. ° C. Decreas

ing the area of the plates. D. Increasing the area of the plates. E. Both A and D F. Both B and C
Physics
1 answer:
eimsori [14]3 years ago
4 0

You can increase the capacitance of a capacitor by decreasing the plate spacing (A) or by increasing the area of the plates (D).

'A' and 'D' both do the job, so the correct choice is<em> (E)</em> .

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The intensity of an earthquake wave passing through the earth is measured to be 2.5×106 j/(m2⋅s) at a distance of 43 km from the
vampirchik [111]

r₁ = distance of the point from the source = 43 km = 43000 m

I₁ = intensity of earthquake wave at distance "r₁" = 2.5 x 10⁶ W/m²

r₂ = distance of the point from the source = 1.5 km = 1500 m

I₂ = intensity of earthquake wave at distance "r₂" = ?

we know that , for a constant power , the intensity of wave is inversely proportional to the distance from the source .

I α 1/r²             where I = intensity of wave , r = distance from source

hence we can write

I₁/I₂ = r₂²/r₁²

inserting the values

(2.5 x 10⁶) /I₂ = (1500/43000)²

I₂ = 2.1 x 10⁹ W/m²

4 0
3 years ago
When a 4-kg ball is thrown upwards at 40 m/s, at what
arlik [135]

Answer:

the height of the potential energy is 3,200 J

Explanation:

The computation of the kinetic energy is shown below:

Kinetic energy = 1 ÷ 2 × mass × velocity^2

= 1 ÷ 2 × 4 kg × 40 m/s^2

= 3,200 J

Hence the height of the potential energy is 3,200 J

4 0
3 years ago
Which waves have wavelengths longer than those of visible light? Give an example of how each kind of wave is used.
Kazeer [188]
1. Radio Waves
ex. Wi-Fi
2. Microwaves
ex. Mobile Phones
3. Infrared Radiation
ex. Heat Lamps
6 0
3 years ago
Read 2 more answers
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IRISSAK [1]

Answer:

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6 0
3 years ago
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A gas is compressed from an initial volume of 5.55 L to a final volume of 1.22 L by an external pressure of 1.00 atm. During the
algol13

Answer:

Explanation:

change in the volume of the gas = 5.55 - 1.22

= 4.33 X 10⁻³ m³

external pressure ( constant ) P = 1 x 10⁵ Pa

work done on the gas

=external pressure x change in volume

= 10⁵ x  4.33 X 10⁻³

=4.33 x 10²

433 J

Using the formula

Q = ΔE + W , Q is heat added , ΔE is change in internal energy , W is work done by the gas

Given

Q = - 124 J ( heat is released so negative )

W = - 433 J . ( work done by gas is negative, because it is done on gas  )

- 124  = ΔE - 433

ΔE = 433  - 124

= 309 J

There is increase of 309 J in the internal energy of the gas.

3 0
3 years ago
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